Thanks FvM for putting up with my silly questions. Your answer makes sense.
Let me be more specific about my problem:
I have a dipole antenna that is impedance matched, so P(radiated)= P(accepted)
That's only true for a loss-less antenna. From a practical point of view, a decent antenna will have high efficiency and radiate most of the RF energy fed to it as RF energy at the same frequency. But some will be converted to heat in the antenna. That can be lost via conduction, convection or radiation (mainly IR)
Note, that the fact the antenna is matched is
not a requirement for that equation to be true. If the antenna is lossless, but poorly matched, it might reflect 40% of the incident power. So if you apply 1 W, 0.4W is reflected, and 0.6 W accepted. The 0.6W will be radiated.
Then, I introduce a lossy dielectric object, consisting mainly from fat and muscle, to the simulated model and put it adjacent to the dipole antenna (5cm away). The simulation results show a decrease in P(radiated) mainly because lossy dielectric object absorbs some of the P(radiated).
Where is there a decrease in power radiated? Are you simply saying the presence of the lossy dielectric causes RF to be reflected by the antenna, so the RF power the antenna radiates is less?
Note the same will happen for a good dielectric like PTFE. Stick a lump of that around an antenna and you will change the resonate frequency. That will cause RF to be reflected from the antenna, and for a constant incident power will mean the antenna radiates less power.
Note: I did not impedance match the antenna in the presence of the lossy dielectric object, only by itself.
Question is: Is it fair to say that P(absorbed by the lossy dielectric object) = P(accepted) - P(radiated)
I disagree with FvM and say that the equation is not true.
As you sated earlier, P(radiated)= P(accepted), which I stated is only true for a lossless antenna. So assuming a lossless antenna, the right hand side of the above equation is zero. The left hand side however is not zero, since the lossy dielectric will heat up. So the equation can't be true. I think the equation you are looking for is:
P(absorbed by the lossy dielectric object) + P(radiated elsewhere) = P(accepted by the antenna) = P(radiated by the lossless antenna)
If the antenna is not lossless, but has an efficiency η, then
P(absorbed by the lossy dielectric object) + P(radiated elsewhere) = P(radiated by the antenna) = η * P(accepted by the antenna)
(When I talk of radiated power, I am assuming you mean immediate RF power - not heat radiated some time later, even after the excitation is removed.)
Now back to my S11 thread: The reason I had a whole post for that is because I read few papers in the literature using S11 measurements to detect the presence of tumors in the breast since tumors have different dielectric properties from fat and more electric field waves would be reflected when tumors are present. My whole understanding of S11 is like what you explained (reflections would be from the antenna structure only), and I am not totally convinced it includes reflections from adjacent lossy dielectric objects, or does it?
Vielen Dank.
The S11 of an antenna in free space is not the same as the S11 you will measure on a network analyzer with items near the antenna. You can measure S11 on many things - mixers, dummy loads etc. If you put something near an antenna, the free space properties of the antenna wont change, but the environment the antenna is in has changed, so S11 will change.
A word of caution about your research, from someone who has spent a long time working in medical physics, although I no longer do.
I used to work in the optics group in the Department of Medical Physics, UCL.
https://www.ucl.ac.uk/medphys/
In fact my PhD is in medical physics, and I was measuring the optical properties of tissue.
**broken link removed**
Some in the group were using the change of scattering coefficient to try to detect tumor tissue. Tumor tissue scatters light more than normal tissue.
https://www.ucl.ac.uk/medphys/research/borl/imaging/monstir/breast
It used to, and looking at the web site, still does, get lots of funding, as one can say you are not using x-rays. But I know spacial resolutions were of the order of cm, and personally I don't feel it is ever going to improve considerably because human tissue is highly scattering to light. The usual technique is to time how long the light takes to get from A to B in the tissue, so then you know how far it has gone. Trouble is, you don't know what path the light took, so building images is very hard.
I was chatting to someone in the X-ray group in the Medical Physics department, who assured me the level of x-rays given for mammograms is very low, but it's possible to detect abnormalities as small as 100 µm. His point was that by the time the optical methods could detect anything reliably, the person (usually female) would be dead. So despite millions of pounds being spent on it, the technique is never going to be of any practical use. I spent more than 10 years working in that group, and believe it will never be of any practical use. It's almost 10 years since I left the group, and the images look no better today than they did 10 years ago.
I rather suspect your RF techniques will suffer the same practical problems as the optical techniques, although for different reasons. Although tumor tissue has different dielectric properties to normal tissue, it is not going to be so dramatically different that you are going to be able to tell them apart with a device suitable for breast screening. If you match your antenna away in free space, and put it near a human breast, there will be a dramatic change in S11. I don't believe you are going to be able to distinguish the types of tissue in such a way to make it practical to produce a clinical instrument on which clinical decisions will be based.
Dave